Solve for $x$ and $y$ using substitution. ${3x+4y = 10}$ ${y = 3x-5}$
Explanation: Since $y$ has already been solved for, substitute $3x-5$ for $y$ in the first equation. ${3x + 4}{(3x-5)}{= 10}$ Simplify and solve for $x$ $3x+12x - 20 = 10$ $15x-20 = 10$ $15x-20{+20} = 10{+20}$ $15x = 30$ $\dfrac{15x}{{15}} = \dfrac{30}{{15}}$ ${x = 2}$ Now that you know ${x = 2}$ , plug it back into $\thinspace {y = 3x-5}\thinspace$ to find $y$ ${y = 3}{(2)}{ - 5}$ $y = 6 - 5$ $y = 1$ You can also plug ${x = 2}$ into $\thinspace {3x+4y = 10}\thinspace$ and get the same answer for $y$ : ${3}{(2)}{ + 4y = 10}$ ${y = 1}$